Finite groups with $\mathfrak{F}$-subnormal normalizers of Sylow subgroups

TL;DR

This paper investigates a class of finite groups where Sylow subgroup normalizers are -subnormal, establishing formation properties and identifying specific hereditary saturated formations where this class coincides with .

Contribution

It introduces and analyzes the class *_ of groups with -subnormal Sylow normalizers, proving it forms a formation and characterizing when it equals certain hereditary saturated formations.

Findings

*_ is a formation.

Identifies hereditary saturated formations where *_ equals .

Provides structural properties of groups with -subnormal Sylow normalizers.

Abstract

Let $\pi$ be a set of primes and $\mathfrak{F}$ be a formation. In this article a properties of the class ${\rm w}^{*}_{\pi}\mathfrak{F}$ of all groups $G$, such that $\pi(G)\subseteq \pi(\mathfrak{F})$ and the normalizers of all Sylow $p$-subgroups of $G$ are $\mathfrak{F}$-subnormal in $G$ for every $p\in\pi\cap\pi(G)$ are investigated. It is established that ${\rm w}^{*}_{\pi}\mathfrak{F}$ is a formation. Some hereditary saturated formations $\mathfrak{F}$ for which ${\rm w}^{*}_{\pi}\mathfrak{F}=\mathfrak{F}$ are founded.

Full text

**Finite groups with $\mathfrak{F}$-subnormal **

normalizers of Sylow subgroups

A. F. Vasil’ev, T. I. Vasil’eva, A. G. Melchenko

Abstract

[TABLE]

Introduction

We consider only finite groups. It is well known what role is played the properties of normalizers of the primary subgroups (local subgroups) in classification of finite simple non-abelian groups. In recent years, local subgroups are actively used in the study of non-simple, in particular, soluble groups. In 1986 it was established [1] that a group is nilpotent if the normalizers of its Sylow subgroups (briefly, Sylow normalizers) are nilpotent. Groups with supersoluble Sylow normalizers were studied in [2-4]. A series of papers [5-9] is dedicated to the study of groups whose all the Sylow normalizers belong to a saturated formation $\mathfrak{F}$.

In this paper, we are interested in the following question. How do the properties of embedding of Sylow normalizers into a group influence on the structure of the whole group?

We note the following results. Group $G$ is nilpotent if and only if its any Sylow normalizer coincide with $G$. By the well-known Glauberman’s theorem [10], if all Sylow subgroups of a group are self-normalizing, then the group is a $p$-group for some prime $p$.

Let $H$ be a subgroup of a group $G$. Consider a chain of subgroups

$H\mathchar 61\relax\nobreak H_{0}\leq H_{1}\leq\dots\leq\ H_{n\mathchar 45\relax\nobreak 1}\leq H_{n}\mathchar 61\relax\nobreak G.\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (1)$

According to [11], $H$ is called $\mathbb{P}$-subnormal in $G$ if either $H\mathchar 61\relax\nobreak G$ or there exists a chain (1) such that $|H_{i}:H_{i\mathchar 45\relax\nobreak 1}|$ is a prime for any $i\mathchar 61\relax\nobreak 1,\dots,n$;

According to [12], $H$ is called ${\rm K}$-$\mathbb{P}$-subnormal in $G$ if there exists a chain (1) such that either $H_{i\mathchar 45\relax\nobreak 1}\unlhd H_{i}$, or $|H_{i}:H_{i\mathchar 45\relax\nobreak 1}|$ is a prime for any $i\mathchar 61\relax\nobreak 1,\ldots\nobreak,n$.

In [13] V.S. Monakhov and V.N. Kniahina established that a group $G$ is supersoluble if and only if all its Sylow normalizers are $\mathbb{P}$-subnormal in $G$.

A subgroup $H$ is called submodular in $G$ [14], if there exists a chain of subgroups (1) such that $H_{i\mathchar 45\relax\nobreak 1}$ is a modular subgroup in $H_{i}$ for $i\mathchar 61\relax\nobreak 1,\ldots\nobreak,s$. Here the modular subgroup in $G$ is a modular element in the lattice of all subgroups of $G$ [15]. The class $s\mathfrak{U}$ of all strongly supersoluble groups was studied in [16] ($s\mathfrak{U}$ is the class of supersoluble groups, in which all Sylow subgroups are submodular). By [17, Theorem 3.2], if the normalizers of all Sylow subgroups of a group $G$ are submodular, then $G\in\nobreak s\mathfrak{U}$.

The concept of subnormality was generalized by T.O. Hawkes [18], L.A. Shemetkov [19] as follows.

Let $\mathfrak{F}$ be a non-empty formation. A subgroup $H$ is called $\mathfrak{F}$-subnormal in $G$ (which is denoted by $H$ $\mathfrak{F}$-${\rm sn}$ $G$), if either $H\mathchar 61\relax\nobreak G$, or there exists a maximal chain (1) such that $H_{i}^{\mathfrak{F}}\leq H_{i\mathchar 45\relax\nobreak 1}$ for $i\mathchar 61\relax\nobreak 1,\dots,n$.

In the case when $\mathfrak{F}$ coincides with the class $\mathfrak{N}$ of all nilpotent groups, every $\mathfrak{N}$-subnormal subgroup is subnormal, the converse is not true in general. However, in soluble groups these concepts are equivalent.

Another generalization of subnormal subgroups was proposed by O. Kegel [21]. We give it according to [20, p. 236].

A subgroup $H$ is called ${\rm K}$-$\mathfrak{F}$-subnormal in $G$ $($which is denoted by $H$ ${\rm K}$-$\mathfrak{F}$-${\rm sn}$ $G)$ if there is a chain of subgroups (1) such either $H_{i\mathchar 45\relax\nobreak 1}\trianglelefteq H_{i}$, or $H_{i}^{\mathfrak{F}}\leq H_{i\mathchar 45\relax\nobreak 1}$ for $i\mathchar 61\relax\nobreak 1,\dots,n$.

Note that a subnormal subgroup is $\rm{K}$-$\mathfrak{F}$-subnormal in any group, the converse is not always true. For the case $\mathfrak{F}\mathchar 61\relax\nobreak\mathfrak{N}$ the concepts of subnormal and $\rm{K}$-$\mathfrak{N}$-subnormal subgroups are equivalent. If $\mathfrak{F}$ coincides with the class $\mathfrak{U}$ of all supersoluble groups, then the concept of $\mathbb{P}$-subnormal subgroup is equivalent to the concept of $\mathfrak{U}$-subnormal and $\rm{K}$-$\mathfrak{U}$-subnormal subgroup in the class of all soluble groups. In an arbitrary group, every $\mathfrak{U}$-subnormal ($\rm{K}$-$\mathfrak{U}$-subnormal) subgroup is $\mathbb{P}$-subnormal ($\rm{K}$-$\mathbb{P}$-subnormal subgroup, respectively), but the converse fails in general.

The monograph [21] reflects the results of many papers in which the properties of $\mathfrak{F}$-subnormal, ${\rm K}$-$\mathfrak{F}$-subnormal subgroups and their applications were studied.

In [22] consideration of the following general problem was started. Let $\mathfrak{F}$ be a non-empty formation. How $\mathfrak{F}$-subnormal $(\rm{K}$-$\mathfrak{F}$-subnormal$)$ Sylow subgroups influence on the structure of the whole group. The classes ${\rm W}_{\pi}\mathfrak{F}$ and $\overline{{\rm W}}_{\pi}\mathfrak{F}$ were investigated in [23]; where ${\rm W}_{\pi}\mathfrak{F}$ ($\overline{{\rm W}}_{\pi}\mathfrak{F}$) is the class of all groups $G$, for which 1 and all Sylow $p$-subgroups are $\mathfrak{F}$-subnormal (respectively $\rm{K}$-$\mathfrak{F}$-subnormal) in $G$ for every $p\in\nobreak\pi\cap\nobreak\pi(G)$. The classes ${\rm W}\mathfrak{F}$ and $\overline{{\rm W}}\mathfrak{F}$ ($\pi$ coincides with the set of all primes) were studied in [24-27]. An interesting generalization of classes ${\rm W}_{\pi}\mathfrak{F}$ and $\overline{{\rm W}}_{\pi}\mathfrak{F}$ was considered in [28].

Definition 1 [29]. *Let $\mathfrak{F}$ be a non-empty formation. A subgroup $H$ of a group $G$ is called strongly ${\rm K}$-$\mathfrak{F}$-subnormal in $G$, if $N_{G}(H)$ is a $\mathfrak{F}$-subnormal subgroup in $G$. *

Note that a subgroup is normal in its normalizer. Therefore every strongly ${\rm K}$-$\mathfrak{F}$-subnormal subgroup is ${\rm K}$-$\mathfrak{F}$-subnormal in any group. The converse is not true. Let $S$ be a symmetric group of degree 3. By [29, theorem B. 10.9] $S$ has an irreducible and faithful $S$-module $U$ over the field $\mathrm{F}_{7}$ of 7 elements. Consider the semidirect product $G\mathchar 61\relax\nobreak[U]S$. The group $G$ is not supersoluble, because $S$ is non-abelian. Since $G/U$ is supersoluble, we see that $H\mathchar 61\relax\nobreak UQ$ is ${\rm K}$-$\mathfrak{U}$-subnormal subgroup of $G$, where $Q$ is a Sylow $3$-subgroup of $G$ that is contained in $S$. Since $H$ is supersoluble, we deduced that $Q$ is ${\rm K}$-$\mathfrak{U}$-subnormal in $G$. Note that the subgroup $Q$ is not strongly ${\rm K}$-$\mathfrak{U}$-subnormal in $G$. This follows from the fact that $N_{G}(Q)\mathchar 61\relax\nobreak S$, but $S$ is not normal and not $\mathfrak{U}$-subnormal in $G$.

Definition 2 [29]. *Given a set of primes $\pi$ and a non-empty formation $\mathfrak{F}$. Introduce the following class of groups: ${\rm w}^{*}_{\pi}\mathfrak{F}$ is the class of all groups $G$, for which $\pi(G)\subseteq\nobreak\pi(\mathfrak{F})$ and all its Sylow $q$-subgroups are strongly $\mathfrak{F}$-subnormal in $G$ for every $q\in\nobreak\pi\cap\nobreak\pi(G)$. *

When $\pi\mathchar 61\relax\nobreak\mathbb{P}$ is the set of all primes, we denote ${\rm w}^{*}_{\mathbb{P}}\mathfrak{F}\mathchar 61\relax\nobreak{\rm w}^{*}\mathfrak{F}$. If $\pi(G)\subseteq\nobreak\pi(\mathfrak{F})$ and $\pi\cap\nobreak\pi(G)\mathchar 61\relax\nobreak\emptyset$, then $N_{G}(1)\mathchar 61\relax\nobreak G$ is $\mathfrak{F}$-subnormal in $G$ and $G\in\nobreak{\rm w}^{*}_{\pi}\mathfrak{F}$.

Problem. Let $\mathfrak{F}$ be a hereditary saturated formation and $\pi$ be some set of primes.

$(1)$* Investigate how the properties of the class ${\rm w}^{*}_{\pi}\mathfrak{F}$ depend on the corresponding properties of $\mathfrak{F}$. In particular, find conditions under which the class ${\rm w}^{*}_{\pi}\mathfrak{F}$ is also a saturated formation;*

$(2)$* Describe $\mathfrak{F}$ for which ${\rm w}^{*}_{\pi}\mathfrak{F}\mathchar 61\relax\nobreak\mathfrak{F}$. *

This paper is devoted studying for some cases of this problem.

1. Preliminary results

We use standard notation and definitions. The appropriate information on groups theory and formations theory can be found in monographs [19], [20] and [30]. We recall some concepts significant in the paper.

By $\mathbb{P}$ we denote the set of all primes. If $\pi\subseteq\nobreak\mathbb{P}$, then $\pi^{\prime}\mathchar 61\relax\nobreak\mathbb{P}\setminus\nobreak\pi$. Let $G$ be a group and $p$ be a prime. We denote by $|G|$ the order of $G$; by $\pi(G)$, the set of all prime divisors of $|G|$; by $O_{p}(G)$, the largest normal $p$-subgroup of $G$; by $O_{\pi}(G)$, the largest normal $\pi$-subgroup of $G$; by ${\rm Syl}_{p}(G)$, the set of all Sylow $p$-subgroups of $G$; by ${\rm Syl}(G)$, the set of all Sylow subgroups of $G$; by $F(G)$, the Fitting subgroup of $G$, which is the largest normal nilpotent subgroup of $G$; by $F_{p}(G)$, the $p$-nilpotent radical of $G$, which is the largest normal $p$-nilpotent subgroup of $G$; by $Z_{p}$, the cyclic group of order $p$; by $1$, the identity subgroup (group).

By $l_{p}(G)$ we denote the $p$-length of the $p$-soluble group $G$; an arithmetic length of the soluble group $G$ is $al(G)\mathchar 61\relax\nobreak\mathrm{Max}\,l_{p}(G)$, where $p$ runs through all primes $p\in\nobreak\pi(G)$; $\mathfrak{L}_{a}(n)$ is the class of all soluble groups $G$ with $al(G)\leq n$; $\mathfrak{L}_{a}(1)$ is the class of all soluble groups $G$ with $al(G)\leq 1$.

In the next lemma, the some familiar properties of Sylow subgroups are collected.

Lemma 1.1. Let $G$ be a group and $p\in\nobreak\mathbb{P}$. Then the following statements are true.

$(1)$* If $P\in\nobreak{\rm Syl}_{p}(G)$ and $N\trianglelefteq G$, then $P\cap\nobreak N\in\nobreak{\rm Syl}_{p}(N)$ and $PN/N\in\nobreak{\rm Syl}_{p}(G/N)$, moreover $N_{G/N}(PN/N)\mathchar 61\relax\nobreak N_{G}(P)N/N$.*

$(2)$* If $H/N\in\nobreak{\rm Syl}_{p}(G/N)$ and $N\trianglelefteq G$, then $H/N\mathchar 61\relax\nobreak PN/N$ for some $P\in\nobreak{\rm Syl}_{p}(G)$.*

$(3)$* If $P\in\nobreak{\rm Syl}(G)$ and $N_{i}\trianglelefteq G$, $i\mathchar 61\relax\nobreak 1,2$, then*

$P\cap\nobreak N_{1}N_{2}\mathchar 61\relax\nobreak(P\cap\nobreak N_{1})(P\cap\nobreak N_{2})$ and $PN_{1}\cap\nobreak PN_{2}\mathchar 61\relax\nobreak P(N_{1}\cap\nobreak N_{2})$.*

$(4)$* If $\pi(G)\mathchar 61\relax\nobreak\{p_{1},\ldots\nobreak,p_{r}\}$ and $P_{i}\in\nobreak{\rm Syl}_{p_{i}}(G)$ for $i\mathchar 61\relax\nobreak 1,\ldots\nobreak,r$, then $G\mathchar 61\relax\nobreak\langle P_{1},\ldots\nobreak,P_{r}\rangle$. *

Lemma 1.2 [30, lemma A.1.2] Let $U$, $V$ and $W$ be subgroups of $G$. Then the following statements are equivalent:

$(1)$* $U\cap\nobreak VW\mathchar 61\relax\nobreak(U\cap\nobreak V)(U\cap\nobreak W);$*

$(2)$* $UV\cap\nobreak UW\mathchar 61\relax\nobreak U(V\cap\nobreak W)$. *

Proposition 1.3. Let $G$ be a group, $P\in\nobreak{\rm Syl}(G)$ and $N_{i}\trianglelefteq G$, $i\mathchar 61\relax\nobreak 1,2$. Then

$N_{G}(P)\cap\nobreak N_{1}N_{2}\mathchar 61\relax\nobreak(N_{G}(P)\cap\nobreak N_{1})(N_{G}(P)\cap\nobreak N_{2})$ and*

$N_{G}(P)N_{1}\cap\nobreak N_{G}(P)N_{2}\mathchar 61\relax\nobreak N_{G}(P)(N_{1}\cap\nobreak N_{2})$.*

Proof. We proceed by induction on $|G|$. Let $N_{1}$ and $N_{2}$ be normal subgroups of $G$ and $P\in\nobreak{\rm Syl}(G)$. If $N_{1}\cap\nobreak N_{2}\neq 1$, then there exist a minimal normal subgroup $N$ of $G$, contained in $N_{1}\cap\nobreak N_{2}$. By induction

$N_{G/N}(PN/N)\cap\nobreak N_{1}/N\cdot\nobreak N_{2}/N\mathchar 61\relax\nobreak(N_{G/N}(PN/N)\cap\nobreak N_{1}/N)(N_{G/N}(PN/N)\cap\nobreak N_{2}/N)$.

By Lemma 1.1(1) $N_{G/N}(PN/N)\mathchar 61\relax\nobreak N_{G}(P)N/N$. By the Dedekind identity, we have $N_{G}(P)N/N\cap\nobreak N_{1}N_{2}/N\mathchar 61\relax\nobreak(N_{G}(P)\cap\nobreak N_{1}N_{2})N/N$ and $N_{G}(P)N/N\cap\nobreak N_{i}/N\mathchar 61\relax\nobreak(N_{G}(P)\cap\nobreak N_{i})N/N$ for $i\mathchar 61\relax\nobreak 1,2$.

Then $N_{G}(P)\cap\nobreak N_{1}N_{2}\mathchar 61\relax\nobreak N_{G}(P)\cap\nobreak(N_{G}(P)N\cap\nobreak N_{1}N_{2})\mathchar 61\relax\nobreak N_{G}(P)\cap\nobreak(N_{G}(P)\cap\nobreak N_{1})N\cdot\nobreak(N_{G}(P)\cap\nobreak N_{2})N\mathchar 61\relax\nobreak(N_{G}(P)\cap\nobreak N_{1})(N_{G}(P)\cap\nobreak N_{2})(N_{G}(P)\cap\nobreak N)\mathchar 61\relax\nobreak(N_{G}(P)\cap\nobreak N_{1})(N_{G}(P)\cap\nobreak N_{2})$.

Let $N_{1}\cap\nobreak N_{2}\mathchar 61\relax\nobreak 1$. Let $T\mathchar 61\relax\nobreak N_{G}(P)N_{1}\cap\nobreak N_{G}(P)N_{2}$. Since $PN_{i}\unlhd N_{G}(P)N_{i}$, $i\mathchar 61\relax\nobreak 1,2$, we have $PN_{1}\cap\nobreak PN_{2}\unlhd T$. From $N_{1}\cap\nobreak N_{2}\mathchar 61\relax\nobreak 1$ and lemma 1.1(3) follows that $PN_{1}\cap\nobreak PN_{2}\mathchar 61\relax\nobreak P(N_{1}\cap\nobreak N_{2})\mathchar 61\relax\nobreak P$. Therefore $P\unlhd T$ and $T\mathchar 61\relax\nobreak N_{G}(P)$. Then $N_{G}(P)(N_{1}\cap\nobreak N_{2})\mathchar 61\relax\nobreak N_{G}(P)\mathchar 61\relax\nobreak N_{G}(P)N_{1}\cap\nobreak N_{G}(P)N_{2}$. By lemma 1.2 $N_{G}(P)\cap\nobreak N_{1}N_{2}\mathchar 61\relax\nobreak(N_{G}(P)\cap\nobreak N_{1})(N_{G}(P)\cap\nobreak N_{2})$. $\Box$

Lemma 1.4. [19, lemma 3.9]. *If $H/K$ is a chief factor of a group $G$ and $p\in\nobreak\pi(H/K)$, then $G/C_{G}(H/K)$ does not contain nonidentity normal $p$-subgroups, and $F_{p}(G)\leq C_{G}(H/K)$. *

Let $\mathfrak{F}$ be a class of groups. By $\pi(\mathfrak{F})$ we denote the set of all prime divisors of orders of groups belonging to $\mathfrak{F}$; $\mathfrak{F}_{\pi}$ is the class of all $\pi$-groups belonging to $\mathfrak{F}$; $\mathfrak{F}_{p}\mathchar 61\relax\nobreak\mathfrak{F}_{\pi}$ for $\pi\mathchar 61\relax\nobreak\{p\}$.

We will use the following notation: $\mathfrak{G}$ is the class of all groups, $\mathfrak{S}$ is the class of all soluble groups, $\mathfrak{N}$ is the class of all nilpotent groups, $\mathfrak{N}^{2}$ is the class of all metanilpotent groups, $\mathfrak{NA}$ is the class of all groups $G$ with the nilpotent commutator subgroup $G^{\prime}$.

A minimal non-$\mathfrak{F}$-group is a group $G$ such that $G\not\in\nobreak\mathfrak{F}$, and any proper subgroup of $G$ belongs to $\mathfrak{F}$. A minimal non-$\mathfrak{N}$-group is called a Schmidt group.

A class of groups $\mathfrak{F}$ is called a formation, if

1. $\mathfrak{F}$ is a homomorph, i.e., from $G\in\nobreak\mathfrak{F}$ and $N\trianglelefteq G$ it follows that $G/N\in\nobreak\mathfrak{F}$ and
2. from $N_{i}\trianglelefteq G$ and $G/N_{i}\in\nobreak\mathfrak{F}$ $(i\mathchar 61\relax\nobreak 1,2)$ it ensues that $G/N_{1}\cap\nobreak N_{2}\in\nobreak\mathfrak{F}$.

A formation $\mathfrak{F}$ is called saturated, if from $G/\Phi(G)\in\nobreak\mathfrak{F}$ it follows that $G\in\nobreak\mathfrak{F}$. A formation $\mathfrak{F}$ is called hereditary if, together with each group, $\mathfrak{F}$ contains all its subgroups. By symbol $G^{\mathfrak{F}}$ denotes the $\mathfrak{F}$-residual of $G$; i.e., the least normal subgroup of $G$ for which $G/G^{\mathfrak{F}}\in\nobreak\mathfrak{F}$.

A function $f:\mathbb{P}\to\{$formations$\}$ is called a local screen. By $LF(f)$ we denote the class of all groups $G$ with $G/C_{G}(H/K)\in\nobreak f(p)$ for each chief factor $H/K$ and each $p\in\nobreak\pi(H/K)$. A formation $\mathfrak{F}$ is called local, if there exists a local screen $f$ with $\mathfrak{F}\mathchar 61\relax\nobreak LF(f)$.

A screen $f$ of a formation $\mathfrak{F}$ is called inner if $f(p)\subseteq\nobreak\mathfrak{F}$ for each prime $p$. An inner screen $f$ of $\mathfrak{F}$ is called the maximal inner if, for its every inner screen $h$, we have $h(p)\subseteq\nobreak f(p)$ for every prime $p$.

Lemma 1.5 [19, lemma 4.5]. *Let $\mathfrak{F}\mathchar 61\relax\nobreak LF(f)$. A group $G$ belongs to $\mathfrak{F}$ if and only if $G/F_{p}(G)\in\nobreak f(p)$ for each $p\in\nobreak\pi(G)$. *

We give some knows properties of $\mathfrak{F}$-subnormal and $\rm{K}$-$\mathfrak{F}$-subnormal subgroups.

Lemma 1.6. Let $\mathfrak{F}$ be a non-empty formation, $H$ and $K$ are subgroups of a group $G$, and $N\trianglelefteq G$.

$(1)$* If $H$ $\mathfrak{F}$-${\rm sn}$ $G$ $(H$ ${\rm K}$-$\mathfrak{F}$-${\rm sn}$ $G)$ then $HN/N$ $\mathfrak{F}$-${\rm sn}$ $G/N$ $(HN/N$ ${\rm K}$-$\mathfrak{F}$-${\rm sn}$ $G/N).$*

$(2)$* If $N\leq H$ and $H/N$ $\mathfrak{F}$-${\rm sn}$ $G/N$ $(H/N$ ${\rm K}$-$\mathfrak{F}$-${\rm sn}$ $G/N)$ then $H$ $\mathfrak{F}$-${\rm sn}$ $G$ $(H$ ${\rm K}$-$\mathfrak{F}$-${\rm sn}$ $G).$*

$(3)$* If $H$ $\mathfrak{F}$-${\rm sn}$ $G$ $(H$ ${\rm K}$-$\mathfrak{F}$-${\rm sn}$ $G)$ then $HN$ $\mathfrak{F}$-${\rm sn}$ $G$ $(HN$ ${\rm K}$-$\mathfrak{F}$-${\rm sn}$ $G).$*

$(4)$* If $H$ $\mathfrak{F}$-${\rm sn}$ $K$ $(H$ ${\rm K}$-$\mathfrak{F}$-${\rm sn}$ $K)$ and $K$ $\mathfrak{F}$-${\rm sn}$ $G$ $(K$ ${\rm K}$-$\mathfrak{F}$-${\rm sn}$ $G)$ then $H$ $\mathfrak{F}$-${\rm sn}$ $G$ $(H$ ${\rm K}$-$\mathfrak{F}$-${\rm sn}$ $G).$*

$(5)$* If all composition factors of $G$ belong to $\mathfrak{F}$ then every subnormal subgroup of $G$ is $\mathfrak{F}$-subnormal.*

$(6)$* Let $p$ be a prime and let $G$ be a $p$-group. If $Z_{p}\in\nobreak\mathfrak{F}$ then all subgroups of $G$ are $\mathfrak{F}$-subnormal. *

Lemma 1.7. Let $\mathfrak{F}$ be a non-empty hereditary formation, $H\leq G$ and $M\leq G$.

$(1)$* If $H$ $\mathfrak{F}$-${\rm sn}$ $G$ $(H$ ${\rm K}$-$\mathfrak{F}$-${\rm sn}$ $G)$ then $H\cap\nobreak M$ $\mathfrak{F}$-${\rm sn}$ $M$ $(H\cap\nobreak M$ ${\rm K}$-$\mathfrak{F}$-${\rm sn}$ $M).$*

$(2)$* If $H$ $\mathfrak{F}$-${\rm sn}$ $G$ and $M$ $\mathfrak{F}$-${\rm sn}$ $G$ $(H$ ${\rm K}$-$\mathfrak{F}$-${\rm sn}$ $G$ and $M$ ${\rm K}$-$\mathfrak{F}$-${\rm sn}$ $G)$ then $H\cap\nobreak M$ $\mathfrak{F}$-${\rm sn}$ $G$ $(H\cap\nobreak M$ ${\rm K}$-$\mathfrak{F}$-${\rm sn}$ $G).$*

$(3)$* If $G^{\mathfrak{F}}\leq H$ then $H$ $\mathfrak{F}$-${\rm sn}$ $G$ $(H$ ${\rm K}$-$\mathfrak{F}$-${\rm sn}$ $G).$*

$(4)$* If $H$ $\mathfrak{F}$-${\rm sn}$ $G$ $(H$ ${\rm K}$-$\mathfrak{F}$-${\rm sn}$ $G)$ then $H^{x}$ $\mathfrak{F}$-${\rm sn}$ $G$ $(H^{x}$ ${\rm K}$-$\mathfrak{F}$-${\rm sn}$ $G)$ for any $x\in\nobreak G$. *

2. Properties of the Class ${\rm w}^{*}_{\pi}\mathfrak{F}$

Recall that the class of groups ${\rm w}^{*}_{\pi}\mathfrak{F}$ is defined as follows:

${\rm w}^{*}_{\pi}\mathfrak{F}\mathchar 61\relax\nobreak(G$ $|$ $\pi(G)\subseteq\nobreak\pi(\mathfrak{F})$ and every Sylow $q$-subgroup of $G$ is strongly $\mathfrak{F}$-subnormal in $G$, where $q\in\nobreak\pi\cap\nobreak\pi(G))$.

The following example shows that ${\rm w}^{*}_{\pi}\mathfrak{F}\not\mathchar 61\relax\nobreak\mathfrak{F}$ in the general case.

Example 2.1. Let $\mathfrak{F}\mathchar 61\relax\nobreak\mathfrak{N}^{3}$ be the formation of all soluble groups whose nilpotent length is $\leq 3$. Take the symmetric group $S_{4}\mathchar 61\relax\nobreak M$ of degree 4. By [30, theorem B. 10.9] there exists an irreducible and faithful $M$-module $U$ over the field $\mathrm{F}_{3}$ of 3 elements. Consider the semidirect product $G\mathchar 61\relax\nobreak[U]M$. Note that the nilpotent length of $G$ is 4 and $\pi(G)\mathchar 61\relax\nobreak\{2,3\}$. Since $S$ is a minimal non-$\mathfrak{N}^{2}$-subgroup, we deduced that $G$ is minimal non-$\mathfrak{N}^{3}$-group. It is easy to see that the normalizers of its Sylow subgroups are $\mathfrak{F}$-subnormal subgroups in $G$, but $G$ does not belong to $\mathfrak{F}$.

Definition 2.2. *A class of groups $\mathfrak{F}$ is called $S_{H}$-closed, if from $G\in\nobreak\mathfrak{F}$ it follows that every Hall subgroup of $G$ belongs to $\mathfrak{F}$. *

Proposition 2.3. Let $\mathfrak{F}$ be a non-empty formation and $\pi\subseteq\nobreak\mathbb{P}$.

$(1)$* If $\pi_{1}$ is a set of primes and $\pi\subseteq\nobreak\pi_{1}$ then ${\rm w}^{*}_{{\pi}_{1}}\mathfrak{F}\subseteq\nobreak{\rm w}^{*}_{\pi}\mathfrak{F}.$*

$(2)$* $\mathfrak{N}_{\pi\cap\nobreak\pi(\mathfrak{F})}\subseteq\nobreak{\rm w}^{*}_{\pi}\mathfrak{F}.$*

$(3)$* ${\rm w}^{*}_{\pi}\mathfrak{F}\mathchar 61\relax\nobreak{\rm w}^{*}_{\pi\cap\nobreak\pi(\mathfrak{F})}\mathfrak{F}.$*

$(4)$* ${\rm w}^{*}_{\pi}\mathfrak{F}$ is a homomorph.*

$(5)$* If a formation $\mathfrak{F}_{1}\subseteq\nobreak\mathfrak{F}$ then ${\rm w}^{*}_{\pi}\mathfrak{F}_{1}\subseteq\nobreak{\rm w}^{*}_{\pi}\mathfrak{F}$.*

Proof. (1): Let $G\in\nobreak{\rm w}^{*}_{{\pi}_{1}}\mathfrak{F}$, $q\in\nobreak\pi\cap\nobreak\pi(G)$ and $Q$ be any Sylow $q$-subgroup of $G$. Since $q\in\nobreak\pi_{1}\cap\nobreak\pi(G)$, we have $N_{G}(Q)\ \mathfrak{F}$-${\rm sn}$ $G$. Hence ${\rm w}^{*}_{{\pi}_{1}}\mathfrak{F}\subseteq\nobreak{\rm w}^{*}_{\pi}\mathfrak{F}$.

(2): Let $G\in\nobreak\mathfrak{N}_{\pi\cap\nobreak\pi(\mathfrak{F})}$. Then $\pi(G)\subseteq\nobreak(\pi\cap\nobreak\pi(\mathfrak{F}))\subseteq\nobreak\pi(\mathfrak{F})$. Since $N_{G}(P)\mathchar 61\relax\nobreak G$ for every $P\in\nobreak{\rm Syl}(G)$, by definition 1 it follows that $G\in\nobreak{\rm w}^{*}_{\pi}\mathfrak{F}$.

(3): From (1) it follows that ${\rm w}^{*}_{\pi}\mathfrak{F}\subseteq\nobreak{{\rm w}^{*}}_{\pi\cap\nobreak\pi(\mathfrak{F})}\mathfrak{F}$.

Let $G\in\nobreak{{\rm w}^{*}}_{\pi\cap\nobreak\pi(\mathfrak{F})}\mathfrak{F}$. Since $\pi(G)\subseteq\nobreak\pi(\mathfrak{F})$, we have $\pi\cap\nobreak\pi(\mathfrak{F})\cap\nobreak\pi(G)\mathchar 61\relax\nobreak\pi\cap\nobreak\pi(G)$. Consequently, if $q\in\nobreak\pi\cap\nobreak\pi(G)$, then in $G$ the normalizer of every Sylow $q$-subgroup is $\mathfrak{F}$-subnormal. So $G\in\nobreak{\rm w}^{*}_{\pi}\mathfrak{F}$ and ${\rm w}^{*}_{\pi}\mathfrak{F}\mathchar 61\relax\nobreak{{\rm w}^{*}}_{\pi\cap\nobreak\pi(\mathfrak{F})}\mathfrak{F}$.

(4): To prove that ${\rm w}^{*}_{\pi}\mathfrak{F}$ is a homomorph, let $G\in\nobreak{\rm w}^{*}_{\pi}\mathfrak{F}$, $N\trianglelefteq G$ and $p\in\nobreak\pi\cap\nobreak\pi(G/N)$. Consider $H/N\in\nobreak{\rm Syl}_{p}(G/N)$. By Lemma 1.1(2) $H/N\mathchar 61\relax\nobreak PN/N$ for some Sylow $p$-subgroup $P$ of $G$. From $G\in\nobreak{\rm w}^{*}_{\pi}\mathfrak{F}$ it follows that $N_{G}(P)$ $\mathfrak{F}$-${\rm sn}$ $G$. Then by Lemma 1.1(1) and Lemma 1.6(1) $N_{G/N}(H/N)\mathchar 61\relax\nobreak N_{G}(P)N/N$ $\mathfrak{F}$-${\rm sn}$ $G/N$. From here and $\pi(G/N)\subseteq\nobreak\pi(G)\subseteq\nobreak\pi(\mathfrak{F})$ we have that $G/N\in\nobreak{\rm w}^{*}_{\pi}\mathfrak{F}$. So ${\rm w}^{*}_{\pi}\mathfrak{F}$ is a homomorph.

(5): Let $G\in\nobreak{\rm w}^{*}_{\pi}\mathfrak{F}_{1}$. Then $\pi(G)\subseteq\nobreak\pi(\mathfrak{F}_{1})\subseteq\nobreak\pi(\mathfrak{F})$. From $q\in\nobreak\pi\cap\nobreak\pi(G)$ it follows that every $Q\in\nobreak{\rm Syl}_{q}(G)$ is strongly ${\rm K}$-$\mathfrak{F}_{1}$-subnormal in $G$. If $N_{G}(Q)\mathchar 61\relax\nobreak G$, then $N_{G}(Q)$ $\mathfrak{F}$-${\rm sn}$ $G$. Suppose that a maximal chain of subgroups $N_{G}(Q)\mathchar 61\relax\nobreak H_{0}\mathchar 60\relax\nobreak H_{1}\mathchar 60\relax\nobreak\dots\mathchar 60\relax\nobreak H_{n}\mathchar 61\relax\nobreak G$ exists and $H_{i}^{\mathfrak{F}_{1}}\leq H_{i\mathchar 45\relax\nobreak 1}$ for $i\mathchar 61\relax\nobreak 1,\dots,n$. From $H_{i}/H_{i}^{\mathfrak{F}_{1}}\in\nobreak\mathfrak{F}_{1}\subseteq\nobreak\mathfrak{F}$ we have $H_{i}^{\mathfrak{F}}\subseteq\nobreak H_{i}^{\mathfrak{F}_{1}}\leq H_{i\mathchar 45\relax\nobreak 1}$. Hence $N_{G}(Q)$ $\mathfrak{F}$-${\rm sn}$ $G$. So ${\rm w}^{*}_{\pi}\mathfrak{F}_{1}\subseteq\nobreak{\rm w}^{*}_{\pi}\mathfrak{F}$. $\Box$

Theorem 2.4. Let $\mathfrak{F}$ be a non-empty hereditary formation and $\pi\subseteq\nobreak\mathbb{P}$. Then

$(1)$* $\mathfrak{F}\subseteq\nobreak{\rm w}^{*}\mathfrak{F}\subseteq\nobreak{\rm w}^{*}_{\pi}\mathfrak{F}$,*

$(2)$* ${\rm w}^{*}_{\pi}\mathfrak{F}$ is an $S_{H}$-closed formation,*

$(3)$* ${\rm w}^{*}_{\pi}\mathfrak{F}\mathchar 61\relax\nobreak{\rm w}^{*}_{\pi}({\rm w}^{*}_{\pi}\mathfrak{F})$.*

Proof. (1): From Lemma 1.7(3) it follows that $\mathfrak{F}\subseteq\nobreak{\rm w}^{*}\mathfrak{F}$. From $\pi\subseteq\nobreak\mathbb{P}$ and Proposition 2.3(1) we conclude that ${\rm w}^{*}\mathfrak{F}\subseteq\nobreak{\rm w}^{*}_{\pi}\mathfrak{F}$.

(2): To prove $S_{H}$-closure of ${\rm w}^{*}_{\pi}\mathfrak{F}$, let $G\in\nobreak{\rm w}^{*}_{\pi}\mathfrak{F}$ and let $H$ be a Hall subgroup of $G$. Then $\pi(H)\subseteq\nobreak\pi(G)\subseteq\nobreak\pi(\mathfrak{F})$. Let $q\in\nobreak\pi\cap\nobreak\pi(H)$ and $S$ be a Sylow $q$-subgroup of $H$. Since $S\in\nobreak{\rm Syl}_{q}(G)$, we have $N_{G}(S)$ $\mathfrak{F}$-${\rm sn}$ $G$. By Lemma 1.7(1) $N_{H}(S)\mathchar 61\relax\nobreak(N_{G}(S)\cap\nobreak H)$ $\mathfrak{F}$-${\rm sn}$ $H$. Therefore $H\in\nobreak{\rm w}^{*}_{\pi}\mathfrak{F}$.

By Proposition 2.3(4) ${\rm w}^{*}_{\pi}\mathfrak{F}$ is a homomorph.

Let us proved that ${\rm w}^{*}_{\pi}\mathfrak{F}$ is closed under subdirect products. Suppose that is false, and let $G$ be a counterexample with $|G|$ as small as possible. Then there exists a subgroup $N_{i}\unlhd G$ such that $G/N_{i}\in\nobreak{{\rm w}^{*}_{\pi}}\mathfrak{F}$, $i\mathchar 61\relax\nobreak 1,2$, but $G/N_{1}\cap\nobreak N_{2}\notin\nobreak{{\rm w}^{*}_{\pi}}\mathfrak{F}$. We note that from $\pi(G/N_{i})\subseteq\nobreak\pi(\mathfrak{F})$, $i\mathchar 61\relax\nobreak 1,2$, it follows that $\pi(G/N_{1}\cap\nobreak N_{2})\subseteq\nobreak\pi(\mathfrak{F})$. By the choice of $G$ we can assume that $N_{1}\cap\nobreak N_{2}\mathchar 61\relax\nobreak 1$. Let $p\in\nobreak\pi\cap\nobreak\pi(G)$ and $R\in\nobreak{\rm Syl}_{p}(G)$. Since $RN_{i}/N_{i}$ is a Sylow $p$-subgroup of $G/N_{i}$ and $G/N_{i}\in\nobreak{{\rm w}^{*}_{\pi}}\mathfrak{F}$, we have $N_{G/N_{i}}(RN_{i}/N_{i})$ $\mathfrak{F}$-${\rm sn}$ $G/N_{i}$, $i\mathchar 61\relax\nobreak 1,2$. By Lemmas 1.1(1) and 1.6(2) $N_{G}(R)N_{i}$ $\mathfrak{F}$-${\rm sn}$ $G$, $i\mathchar 61\relax\nobreak 1,2$. From Lemma 1.7(2) it follows $N_{G}(R)N_{1}\cap\nobreak N_{G}(R)N_{2}$ $\mathfrak{F}$-${\rm sn}$ $G$. From Proposition 1.3 we conclude that $N_{G}(R)N_{1}\cap\nobreak N_{G}(R)N_{2}\mathchar 61\relax\nobreak N_{G}(R)(N_{1}\cap\nobreak N_{2})\mathchar 61\relax\nobreak N_{G}(R)$ $\mathfrak{F}$-${\rm sn}$ $G$. We have the contradiction to the choice of $G$. So ${\rm w}^{*}_{\pi}\mathfrak{F}$ is closed under subdirect products.

(3): Denote $\mathfrak{X}\mathchar 61\relax\nobreak{{\rm w}^{*}_{\pi}}\mathfrak{F}$. Let $G\in\nobreak\mathfrak{X}$. Then $\pi(G)\subseteq\nobreak\pi(\mathfrak{F})$. By (1) we have that $\mathfrak{F}\subseteq\nobreak\mathfrak{X}$. Therefore $\pi(G)\subseteq\nobreak\pi(\mathfrak{X})$. Let $q\in\nobreak\pi\cap\nobreak\pi(G)$ and $Q\in\nobreak{\rm Syl}_{q}(G)$. From $G\in\nobreak\mathfrak{X}$ it follows that $N_{G}(Q)$ $\mathfrak{F}$-${\rm sn}$ $G$. Assume that $N_{G}(Q)\not\mathchar 61\relax\nobreak G$. Then there is a maximal chain of subgroups $N_{G}(Q)\mathchar 61\relax\nobreak H_{0}\mathchar 60\relax\nobreak H_{1}\mathchar 60\relax\nobreak\dots\mathchar 60\relax\nobreak H_{n}\mathchar 61\relax\nobreak G$ such that $H_{i}^{\mathfrak{F}}\leq H_{i\mathchar 45\relax\nobreak 1}$ for $i\mathchar 61\relax\nobreak 1,\dots,n$. By (2) $\mathfrak{X}$ is a formation. Therefore from $H_{i}/H_{i}^{\mathfrak{F}}\in\nobreak\mathfrak{F}\subseteq\nobreak\mathfrak{X}$ it follows that $H_{i}^{\mathfrak{X}}\leq H_{i}^{\mathfrak{F}}\leq H_{i\mathchar 45\relax\nobreak 1}$. This means that $N_{G}(Q)$ $\mathfrak{X}$-${\rm sn}$ $G$. If $N_{G}(Q)\mathchar 61\relax\nobreak G$, then $N_{G}(Q)$ $\mathfrak{X}$-${\rm sn}$ $G$. So $G\in\nobreak{{\rm w}^{*}_{\pi}}\mathfrak{X}$ and $\mathfrak{X}\subseteq\nobreak{{\rm w}^{*}_{\pi}}\mathfrak{X}$ is proved.

Suppose that $\mathfrak{X}\not\mathchar 61\relax\nobreak{{\rm w}^{*}_{\pi}}\mathfrak{X}$. Let $G$ be the group of minimal order in ${{\rm w}^{*}_{\pi}}\mathfrak{X}\textbackslash\mathfrak{X}$. Then $\pi(G)\subseteq\nobreak\pi(\mathfrak{X})\subseteq\nobreak\pi(\mathfrak{F})$. Since $G\not\in\nobreak\mathfrak{X}$, there exists $P\in\nobreak{\rm Syl}_{p}(G)$ such that $p\in\nobreak\pi\cap\nobreak\pi(G)$ and $N_{G}(P)$ is not $\mathfrak{F}$-subnormal in $G$. We note that $N_{G}(P)$ $\mathfrak{X}$-${\rm sn}$ $G$. Then $N_{G}(P)\not\mathchar 61\relax\nobreak G$ and there exists a maximal chain of subgroups $N_{G}(P)\mathchar 61\relax\nobreak H_{0}\mathchar 60\relax\nobreak H_{1}\mathchar 60\relax\nobreak\dots\mathchar 60\relax\nobreak H_{n\mathchar 45\relax\nobreak 1}\mathchar 60\relax\nobreak H_{n}\mathchar 61\relax\nobreak G$ such that $H_{i}^{\mathfrak{X}}\leq H_{i\mathchar 45\relax\nobreak 1}$ for $i\mathchar 61\relax\nobreak 1,\dots,n$. Since $N_{G}(P)\mathchar 61\relax\nobreak N_{H_{i}}(P)$, $N_{H_{i}}(P)H_{i}^{\mathfrak{X}}\leq H_{i\mathchar 45\relax\nobreak 1}$ and $H_{i}/H_{i}^{\mathfrak{X}}\in\nobreak\mathfrak{X}$, we have $N_{H_{i}}(P)H_{i}^{\mathfrak{X}}/H_{i}^{\mathfrak{X}}\mathchar 61\relax\nobreak N_{H_{i}/H_{i}^{\mathfrak{X}}}(PH_{i}^{\mathfrak{X}}/H_{i}^{\mathfrak{X}})$ $\mathfrak{F}$-${\rm sn}$ $H_{i}/H_{i}^{\mathfrak{X}}$. By Lemma 1.6(2) $N_{H_{i}}(P)H_{i}^{\mathfrak{X}}$ $\mathfrak{F}$-${\rm sn}$ $H_{i}$ for $i\mathchar 61\relax\nobreak 1,\dots,n$. Therefore $H_{n}^{\mathfrak{X}}\mathchar 61\relax\nobreak G^{\mathfrak{X}}\not\subseteq\nobreak N_{G}(P)$. From the maximality of $N_{G}(P)$ in $H_{1}$ it follows that $N_{G}(P)$ $\mathfrak{F}$-${\rm sn}$ $H_{1}$. So $n\not\mathchar 61\relax\nobreak 1$. Suppose that $n\mathchar 61\relax\nobreak 2$. Then by Lemma 1.7(1) $N_{G}(P)\mathchar 61\relax\nobreak N_{G}(P)\cap\nobreak N_{G}(P)H_{2}^{\mathfrak{X}}$ $\mathfrak{F}$-${\rm sn}$ $N_{G}(P)H_{2}^{\mathfrak{X}}$. From $N_{G}(P)H_{2}^{\mathfrak{X}}$ $\mathfrak{F}$-${\rm sn}$ $H_{2}$ we conclude that $N_{G}(P)$ $\mathfrak{F}$-${\rm sn}$ $H_{2}\mathchar 61\relax\nobreak G$. This is the contradiction with the choice of $G$. So, we can assume that $n\geq 3$ and $N_{G}(P)$ $\mathfrak{F}$-${\rm sn}$ $H_{n\mathchar 45\relax\nobreak 1}$. Since $N_{G}(P)H_{n}^{\mathfrak{X}}\leq H_{n\mathchar 45\relax\nobreak 1}$, by Lemma 1.7(1) we have $N_{G}(P)\mathchar 61\relax\nobreak N_{G}(P)\cap\nobreak N_{G}(P)H_{n}^{\mathfrak{X}}$ $\mathfrak{F}$-${\rm sn}$ $N_{G}(P)H_{n}^{\mathfrak{X}}$. From $N_{G}(P)H_{n}^{\mathfrak{X}}$ $\mathfrak{F}$-${\rm sn}$ $G$ it follows that $N_{G}(P)$ $\mathfrak{F}$-${\rm sn}$ $G$. This contradicts the choice of $G$. So $\mathfrak{X}\mathchar 61\relax\nobreak{{\rm w}^{*}_{\pi}}\mathfrak{X}$. $\Box$

3. Formations $\mathfrak{F}$ for which ${\rm w}^{*}_{\pi}\mathfrak{F}\mathchar 61\relax\nobreak\mathfrak{F}$

This section focuses on (2) of Problem.

Lemma 3.1. $(1)$* The class $\mathfrak{L}_{a}(1)$ is a hereditary saturated Fitting formation.*

$(2)$* Let $G$ be a soluble group, $\Phi(G)\mathchar 61\relax\nobreak 1$. $G$ is a minimal non-$\mathfrak{L}_{a}(1)$-group if and only if the following statements hold:*

$1)$* $|G|\mathchar 61\relax\nobreak p^{\alpha}q^{\beta}$, $l_{p}(G)\mathchar 61\relax\nobreak 1$, $l_{q}(G)\mathchar 61\relax\nobreak 2$, $l(G)\mathchar 61\relax\nobreak 3$;*

$2)$* $G$ has precisely three conjugate classes of maximal subgroups, whose representatives have the following structure: $G_{p}\leftthreetimes\nobreak G_{q}^{*}$, the Schmidt group, $F(G)\leftthreetimes\nobreak G_{p}$ and $G_{q}\leftthreetimes\nobreak\Phi(G_{p})$, where $G_{q}\mathchar 61\relax\nobreak F(G)\leftthreetimes\nobreak G_{q}^{*}$.*

Proof. (1): The statement follows directly from the fact that $\mathfrak{L}_{a}(1)\mathchar 61\relax\nobreak\cap\nobreak\mathfrak{G}_{p^{\prime}}\mathfrak{N}_{p}\mathfrak{G}_{p^{\prime}}$ for all $p\in\nobreak\mathbb{P}$.

(2): The statement is Lemma 4.1 in [31]. $\Box$

Lemma 3.2. Let $G$ be a biprimary group and let $G\in\nobreak\mathfrak{L}_{a}(1)$. Then $G$ is metanilpotent.

Proof. Let $G$ be a counterexample of minimal order to the statement of the lemma. Since $\mathfrak{N}^{2}$ is a hereditary saturated formation, the group $G\mathchar 61\relax\nobreak NM$, where $N$ is a unique minimal normal subgroup of $G$ and $M$ is a maximal subgroup of $G$, moreover, $N$ is an abelian $p$-group, $p$ is some prime, $M$ is a Schmidt group with a normal $p$-subgroup. From $O_{p}(M)\mathchar 61\relax\nobreak 1$ we conclude that $p$-length of $G$ is 2. This contradicts the fact that $G\in\nobreak\mathfrak{L}_{a}(1)$. $\Box$

Lemma 3.3. Let $\mathfrak{F}$ be a non-empty hereditary formation and let $G$ be a soluble group. If $G\in\nobreak\mathfrak{L}_{a}(1)$, $G\not\mathchar 61\relax\nobreak N_{G}(P)$ and $N_{G}(P)\in\nobreak\mathfrak{F}$ for all $P\in\nobreak{\rm Syl}(G)$, then $G\in\nobreak\mathfrak{F}$.

Proof. Let $G$ be a counterexample of minimal order to the statement of the lemma. Let $N$ is a minimal normal subgroup of $G$. We will prove that $G/N\in\nobreak\mathfrak{F}$. If $G/N\not\mathchar 61\relax\nobreak N_{G/N}(H/N)$ for all $H/N\in\nobreak{\rm Syl}(G/N)$, then $G/N\in\nobreak\mathfrak{F}$ by the choice of $G$. If $G/N\mathchar 61\relax\nobreak N_{G/N}(H/N)$ for some $H/N\in\nobreak{\rm Syl}_{q}(G/N)$, then $H/N\mathchar 61\relax\nobreak QN/N$ for some $Q\in\nobreak{\rm Syl}_{q}(G)$ and $G\mathchar 61\relax\nobreak N_{G}(Q)N$. Since $G$ is solvable and $G\not\mathchar 61\relax\nobreak N_{G}(Q)$, we conclude that $N_{G}(Q)$ is a maximal subgroup of $G$ and $N_{G}(Q)\cap\nobreak N\mathchar 61\relax\nobreak 1$. From here $G/N\cong\nobreak N_{G}(Q)\in\nobreak\mathfrak{F}$. If $K$ is a minimal normal subgroup of $G$ and $K\not\mathchar 61\relax\nobreak N$, then $G/K\in\nobreak\mathfrak{F}$. Since $\mathfrak{F}$ is a formation, we deduce that $G/N\cap\nobreak K\cong\nobreak G\in\nobreak\mathfrak{F}$. This contradicts to the choice of $G$. Consequently $N$ is the unique minimal normal subgroup of $G$. Since $G$ is soluble, we conclude that $N$ is a $p$-group. From the uniqueness of $N$ it follows that $F(G)$ is a $p$-group. By the choice of $G$ imply that $\pi(G)\geq 2$. If $F(G)\not\mathchar 61\relax\nobreak P\in\nobreak{\rm Syl}_{p}(G)$, then $p\in\nobreak\pi(G/F(G))$. This contradicts with $G\in\nobreak\mathfrak{L}_{a}(1)$. Therefore, $F(G)\mathchar 61\relax\nobreak P\in\nobreak{\rm Syl}_{p}(G)$ and $G\mathchar 61\relax\nobreak N_{G}(P)$. This contradiction completes the proof of the lemma. $\Box$

Theorem 3.4. *Let $\mathfrak{F}$ be a hereditary saturated formation and $\mathfrak{F}\subseteq\nobreak\mathfrak{L}_{a}(1)$. A group $G\in\nobreak\mathfrak{F}$ if and only if $\pi(G)\subseteq\nobreak\pi(\mathfrak{F})$ and all its Sylow subgroups are strongly ${\rm K}$-$\mathfrak{F}$-subnormal in $G$. *

Proof. Necessity. Let $G\in\nobreak\mathfrak{F}$. By Lemma 1.7(3) $N_{G}(S)$ $\mathfrak{F}$-${\rm sn}$ $G$ for any Sylow subgroup $S$ of $G$.

Sufficiency. Let $G$ be a counterexample of minimal order and let $N$ be a minimal normal subgroup of $G$.

If $G\mathchar 61\relax\nobreak N$ then $G$ is a simple group, because $N$ is the minimal normal subgroup of $G$. If $G\cong\nobreak Z_{p}$ then from $\pi(G)\subseteq\nobreak\pi(\mathfrak{F})$ it follows that $G\in\nobreak\mathfrak{F}$. This is the contradiction to the choice of $G$. Suppose $G$ is a simple non-abelian group and $p\in\nobreak\pi(G)$. Let $G_{p}\in\nobreak{\rm Syl}_{p}(G)$. Then $N_{G}(G_{p})\neq G$. From $G\notin\nobreak\mathfrak{F}$ it follows that $G^{\mathfrak{F}}\mathchar 61\relax\nobreak G$. By hypothesis $N_{G}(G_{p})$ $\mathfrak{F}$-${\rm sn}$ $G$. Then there is a maximal subgroup $M$ of $G$ such that $N_{G}(G_{p})\subseteq\nobreak M$ and $G^{\mathfrak{F}}\subseteq\nobreak M$. This is the contradiction with $G^{\mathfrak{F}}\mathchar 61\relax\nobreak G$.

Let $N\neq G$. From (1)–(2) of Lemma 1.1, (1) of Lemma 1.6 and hypothesis we have $N_{G/N}(H/N)$ $\mathfrak{F}$-$\mathrm{sn}$ $G/N$ for all $H/N\in\nobreak{\rm Syl}_{q}(G/N)$. By the choice of $G$ we obtain that $G/N\in\nobreak\mathfrak{F}$. If $K$ is a minimal normal subgroup of $G$ and $K\neq N$, then $G/K\in\nobreak\mathfrak{F}$. Since $\mathfrak{F}$ is a formation, we conclude that $G/N\cap\nobreak K\cong\nobreak G\in\nobreak\mathfrak{F}$. This is the contradiction with the choice of $G$. Hence $G$ has the unique minimal normal subgroup $N$. If $\Phi(G)\neq 1$, then from $G/\Phi(G)\in\nobreak\mathfrak{F}$ and saturation $\mathfrak{F}$ it follows that $G\in\nobreak\mathfrak{F}$. This contradicts our assumption. Therefore $\Phi(G)\mathchar 61\relax\nobreak 1$. In this case $N\mathchar 61\relax\nobreak G^{\mathfrak{F}}$ and there is a maximal subgroup $M$ in $G$ such that $G\mathchar 61\relax\nobreak NM$. Consider the following cases.

1. $N$ is a non-abelian group. Let $p\in\nobreak\pi(N)$ and let $G_{p}\in\nobreak{\rm Syl}_{p}(G)$. Then $N_{G}(G_{p})\neq G$. Otherwise $G_{p}\unlhd G$ and $N\subseteq\nobreak G_{p}$, since $N$ is the unique minimal normal subgroup of $G$. But then $N$ is an abelian group. This is contradiction with the proposition.

Consider $N_{G}(G_{p})N$. Let $N_{G}(G_{p})N\mathchar 61\relax\nobreak G$. From $N_{G}(G_{p})$ $\mathfrak{F}$-$\mathrm{sn}$ $G$ we deduce that there is a maximal subgroup $W$ of $G$ such that $N_{G}(G_{p})\subseteq\nobreak W$ and $N\mathchar 61\relax\nobreak G^{\mathfrak{F}}\subseteq\nobreak W$. So we have the contradiction $G\mathchar 61\relax\nobreak N_{G}(G_{p})N\subseteq\nobreak W\neq G$.

Now let $N_{G}(G_{p})N\neq G$. Note that $G_{p}\cap\nobreak N\mathchar 61\relax\nobreak N_{p}\in\nobreak{\rm Syl}_{p}(N)$ and $N_{p}\mathchar 61\relax\nobreak G_{p}\cap\nobreak N\unlhd N_{G}(G_{p})\cap\nobreak N$. Since $N_{G}(G_{p})$ $\mathfrak{F}$-$\mathrm{sn}$ $G$, we see that ($N_{G}(G_{p})\cap\nobreak N$) $\mathfrak{F}$-$\mathrm{sn}$ $N$ by Lemma 1.7(1). Since $N$ is a minimal normal subgroup of $G$, we have either $N^{\mathfrak{F}}\mathchar 61\relax\nobreak 1$ or $N^{\mathfrak{F}}\mathchar 61\relax\nobreak N$. The case $N^{\mathfrak{F}}\mathchar 61\relax\nobreak 1$ is impossible, since $N$ is non-abelian, and $\mathfrak{F}\subseteq\nobreak\mathfrak{S}$. Therefore $N^{\mathfrak{F}}\mathchar 61\relax\nobreak N$. By [30, proposition A.4.13(a)] $N$ is a direct product of subgroups, each isomorphic with a fixed simple non-abelian group. If $N_{G}(G_{p})\cap\nobreak N\mathchar 61\relax\nobreak N$, then $N_{p}\mathchar 61\relax\nobreak G_{p}\cap\nobreak N\unlhd N$. By [30, proposition A.4.13(b)] $N_{p}$ is the direct product of a subset of the non-abelian factors of $N$. This is the contradiction. If $N_{G}(G_{p})\cap\nobreak N\not\mathchar 61\relax\nobreak N$, then there is maximal subgroup $M$ of $N$ such that $N_{G}(G_{p})\cap\nobreak N\leq M$ and $N^{\mathfrak{F}}\leq M$. We have the contradiction $N\mathchar 61\relax\nobreak N^{\mathfrak{F}}\leq M\not\mathchar 61\relax\nobreak N$.

2. $N$ is an abelian $p$-group, $p$ is some prime. From $G/N\in\nobreak\mathfrak{F}\subseteq\nobreak\mathfrak{S}$ and $N\in\nobreak\mathfrak{S}$ it follows that $G$ is solvable. From the uniqueness of $N$ and $\Phi(G)\mathchar 61\relax\nobreak 1$ we conclude that $G\mathchar 61\relax\nobreak N\leftthreetimes\nobreak M$, where $G^{\mathfrak{F}}\mathchar 61\relax\nobreak N\mathchar 61\relax\nobreak C_{G}(N)\mathchar 61\relax\nobreak F(G)$ and $M$ is a maximal subgroup of $G$, and moreover, $M\in\nobreak\mathfrak{F}\subseteq\nobreak\mathfrak{L}_{a}(1)$.

Suppose that $M$ is nilpotent. By Lemma 1.4 $O_{p}(M)\mathchar 61\relax\nobreak 1$, therefore $p\cap\nobreak\pi(M)\mathchar 61\relax\nobreak\emptyset$. It follows that $M$ contains a normal Sylow $q$-subgroup $M_{q}$ for some $q\in\nobreak\pi(M)$ and $q\neq p$. Therefore $M_{q}\mathchar 61\relax\nobreak G_{q}$ is a Sylow $q$-subgroup of the group $G$. From the uniqueness of $N$ it follows that $N_{G}(G_{q})\neq G$. Since $M$ is a maximal subgroup of $G$ and $M\subseteq\nobreak N_{G}(G_{q})$, we have $M\mathchar 61\relax\nobreak N_{G}(G_{q})$. But this contradicts the fact that $N_{G}(G_{q})$ $\mathfrak{F}$-$\mathrm{sn}$ $G$.

We assume that $M$ is non-nilpotent. Let $\pi(G)\mathchar 61\relax\nobreak\{p_{1},p_{2},\ldots\nobreak,p_{n}\}$, where $p_{1}\mathchar 61\relax\nobreak p$. Consider the following cases.

i) Let $n\mathchar 61\relax\nobreak 2$. Then $p\in\nobreak\pi(M)$. By Lemma 1.4 $O_{p}(M)\mathchar 61\relax\nobreak 1$. Since $M\in\nobreak\mathfrak{L}_{a}(1)$, by Lemma 3.2 $M\in\nobreak\mathfrak{N}^{2}$. Therefore $M/F(M)$ is nilpotent. We note that $F(M)$ is a $p_{2}$-group. If $Q\in\nobreak{\rm Syl}_{p_{2}}(M)$, then $Q$ is a normal subgroup of $M$, moreover, $Q\in\nobreak{\rm Syl}_{p_{2}}(G)$ and $N_{G}(Q)\mathchar 61\relax\nobreak M$. By hypothesis $N_{G}(Q)\mathchar 61\relax\nobreak M$ $\mathfrak{F}$-$\mathrm{sn}$ $G$. Therefore $N\mathchar 61\relax\nobreak G^{\mathfrak{F}}\subseteq\nobreak M$ and $G\mathchar 61\relax\nobreak NM\subseteq\nobreak M$. This is the contradiction.

ii) Let $n\geq 3$.

We will to show that $N$ is a Sylow $p$-subgroup of $G$. By Hall’s theorem $G\mathchar 61\relax\nobreak G_{1}G_{2}\cdots G_{n}$, where $G_{1}$, $G_{2},\ldots\nobreak,G_{n}$ are pairwise permutable Sylow $p_{1}$-, $p_{2}$-, $\ldots\nobreak$ , $p_{n}$-subgroups of $G$, respectively. Let $A_{i}\mathchar 61\relax\nobreak G_{1}G_{i}$, where $i\not\mathchar 61\relax\nobreak 1$. Since $|A_{i}|\mathchar 60\relax\nobreak|G|$, $N_{G}(G_{1})\cap\nobreak A_{i}\mathchar 61\relax\nobreak N_{A_{i}}(G_{1})$ $\mathfrak{F}$-$\mathrm{sn}$ $A_{i}$ and $N_{G}(G_{i})\cap\nobreak A_{i}\mathchar 61\relax\nobreak N_{A_{i}}(G_{i})$ $\mathfrak{F}$-$\mathrm{sn}$ $A_{i}$, we have $A_{i}\in\nobreak\mathfrak{F}$. From $|\pi(A_{i})|\mathchar 61\relax\nobreak 2$ by Lemma 3.2 it follows that $A_{i}\in\nobreak\mathfrak{N}^{2}$. We note that $N\subseteq\nobreak A_{i}$. Since $N\mathchar 61\relax\nobreak C_{G}(N)$ and $p_{1}\mathchar 61\relax\nobreak p$, we see that $F(A_{i})$ is a $p$-group. From $A_{i}\in\nobreak\mathfrak{N}^{2}$ it follows that $A_{i}/F(A_{i})\in\nobreak\mathfrak{N}$. Then $G_{1}/F(A_{i})\unlhd A_{i}/F(A_{i})$ and $G_{1}\unlhd A_{i}$. So, $G_{i}\subseteq\nobreak A_{i}\subseteq\nobreak N_{G}(G_{1})$. Hence $G\subseteq\nobreak N_{G}(G_{1})$ and $G_{1}\unlhd G$. From $G_{1}\cap\nobreak M\unlhd M$ and $O_{p}(M)\mathchar 61\relax\nobreak 1$ it follows that $G_{1}\cap\nobreak M\mathchar 61\relax\nobreak 1$. So $G_{1}\mathchar 61\relax\nobreak N\in\nobreak{\rm Syl}_{p}(G)$.

Thus $M$ is a $p^{\prime}$-Hall subgroup of $G$. Let $i\in\nobreak\{2,\ldots\nobreak,n\}$ and $S\in\nobreak{\rm Syl}_{p_{i}}(M)$. Then $S\in\nobreak{\rm Syl}_{p_{i}}(G)$ and $N_{G}(S)\neq M$. We note that $N_{G}(S)\neq G$ because $N\mathchar 61\relax\nobreak C_{G}(N)$ and $N$ is a $p$-group, $p\neq p_{i}$.

We will to show that $N_{G}(S)\in\nobreak\mathfrak{F}$.

Suppose that $N_{G}(S)\cap\nobreak N\mathchar 61\relax\nobreak 1$. Since $G/N\in\nobreak\mathfrak{F}$ and $\mathfrak{F}$ is a hereditary formation, it follows that $N_{G}(S)N/N\cong\nobreak N_{G}(S)/N_{G}(S)\cap\nobreak N\cong\nobreak N_{G}(S)\in\nobreak\mathfrak{F}$.

Suppose now that $N_{G}(S)\cap\nobreak N\mathchar 61\relax\nobreak D\neq 1$. Then $D\unlhd N_{G}(S)$ and $S\unlhd N_{G}(S)$. We have $S\times\nobreak D\unlhd N_{G}(S)$ and $N_{G}(S)\mathchar 61\relax\nobreak(S\times\nobreak D)\leftthreetimes\nobreak R$, where $R$ is a $\{p_{1},p_{i}\}^{\prime}$-Hall subgroup of $N_{G}(S)$. From $G\in\nobreak\mathfrak{S}$ by Hall’s theorem we deduce that $SR\leq M^{x}$ for some $x\in\nobreak G$ and there is a $\{p_{i}\}^{\prime}$-Hall subgroup $H$ from $G$ such that $DR\leq H$. From ${\rm Syl}(H)\subseteq\nobreak{\rm Syl}(G)$ it follows that $N_{G}(L)$ $\mathfrak{F}$-$\mathrm{sn}$ $G$ for any $L\in\nobreak{\rm Syl}(H)$. By Lemma 1.7(1) $N_{H}(L)\mathchar 61\relax\nobreak N_{G}(L)\cap\nobreak H$ $\mathfrak{F}$-$\mathrm{sn}$ $H$. Then $H\in\nobreak\mathfrak{F}$ by the choice of $G$. We note that $M^{x}\cong\nobreak M\in\nobreak\mathfrak{F}$. Since $\mathfrak{F}$ is hereditary we have $N_{G}(S)/D\cong\nobreak SR\in\nobreak\mathfrak{F}$ and $N_{G}(S)/S\cong\nobreak DR\in\nobreak\mathfrak{F}$. We obtain $N_{G}(S)/S\cap\nobreak D\cong\nobreak N_{G}(S)\in\nobreak\mathfrak{F}$.

Consider $T\mathchar 61\relax\nobreak NN_{G}(S)$. From Lemma 1.7(1) $N_{G}(S)$ $\mathfrak{F}$-$\mathrm{sn}$ $T$. By theorem 15.10 [19] $T\in\nobreak\mathfrak{F}$. Let $h$ be the maximal inner local screen formation $\mathfrak{F}$. By Lemma 1.5 [19] it follows that $T/F_{p}(T)\in\nobreak h(p)$. Because $N\leq F_{p}(T)$ and $N\mathchar 61\relax\nobreak C_{G}(N)$, we have $O_{p^{\prime}}(T)\mathchar 61\relax\nobreak 1$ and $N\mathchar 61\relax\nobreak F_{p}(T)$. Therefore $T/N\in\nobreak h(p)$. Then $N_{G}(S)N/N\cong\nobreak N_{G}(S)/N_{G}(S)\cap\nobreak N\in\nobreak h(p)$. Since $\mathfrak{F}$ is a hereditary formation, it follows that $h(p)$ is a hereditary formation, by the theorem 4.7 [19]. Then $(N_{G}(S)\cap\nobreak M)N/N\cong\nobreak N_{G}(S)\cap\nobreak M/N_{G}(S)\cap\nobreak N\cap\nobreak M\cong\nobreak N_{G}(S)\cap\nobreak M\in\nobreak h(p)$. We note that $N_{G}(S)\cap\nobreak M\mathchar 61\relax\nobreak N_{M}(S)$. Therefore $N_{M}(S)\in\nobreak h(p)$. By Lemma 3.4 $M\in\nobreak h(p)$. Then $G/F_{p}(G)\cong\nobreak M\in\nobreak h(p)$. By Lemma 1.5 $G\in\nobreak\mathfrak{F}$, which contradicts the choice of $G$. $\Box$

Corollary 3.4.1 [13]. If the normalizers of all Sylow subgroups of a group $G$ are $\mathbb{P}$-subnormal, then $G$ is supersoluble.

Corollary 3.4.2 [29]. A group $G\in\nobreak\mathfrak{N}^{2}$ if and only if all its Sylow subgroups are strongly ${\rm K}$-$\mathfrak{N}^{2}$-subnormal in $G$.

Corollary 3.4.3 [29]. A group $G\in\nobreak\mathfrak{NA}$ if and only if all its Sylow subgroups are strongly ${\rm K}$-$\mathfrak{NA}$-subnormal in $G$.

Corollary 3.4.4. A group $G\in\nobreak\mathfrak{L}_{a}(1)$ if and only if all its Sylow subgroups are strongly ${\rm K}$-$\mathfrak{L}_{a}(1)$-subnormal in $G$.

Remark 3.5. Note that ${\rm w}^{*}_{\pi}\mathfrak{F}\subseteq\nobreak\overline{{\rm W}}_{\pi}\mathfrak{F}$. From [23, 25] it follows that $\overline{{\rm W}}\mathfrak{N}^{2}\mathchar 61\relax\nobreak{\rm w}\mathfrak{N}^{2}\mathchar 61\relax\nobreak\mathfrak{S}$. But ${\rm w}^{*}\mathfrak{N}^{2}\mathchar 61\relax\nobreak\mathfrak{N}^{2}$.

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A. F. Vasil’ev, A. G. Melchenko

Francisk Skorina Gomel State University, Gomel, Belarus.

E-mail address: formation56@mail.ru, melchenkonastya@mail.ru

Т. I. Vasil’eva

Francisk Skorina Gomel State University, Belarusian State University of Transport, Gomel, Belarus.

E-mail address: tivasilyeva@mail.ru